Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $r \neq 0$. $q = \dfrac{r^2 - 6r - 16}{r^2 - 4r - 12} \times \dfrac{r - 6}{-3r - 9} $
First factor out any common factors. $q = \dfrac{r^2 - 6r - 16}{r^2 - 4r - 12} \times \dfrac{r - 6}{-3(r + 3)} $ Then factor the quadratic expressions. $q = \dfrac {(r + 2)(r - 8)} {(r + 2)(r - 6)} \times \dfrac {r - 6} {-3(r + 3)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac { (r + 2)(r - 8) \times (r - 6)} { (r + 2)(r - 6) \times -3(r + 3)} $ $q = \dfrac {(r + 2)(r - 8)(r - 6)} {-3(r + 2)(r - 6)(r + 3)} $ Notice that $(r + 2)$ and $(r - 6)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac {\cancel{(r + 2)}(r - 8)(r - 6)} {-3\cancel{(r + 2)}(r - 6)(r + 3)} $ We are dividing by $r + 2$ , so $r + 2 \neq 0$ Therefore, $r \neq -2$ $q = \dfrac {\cancel{(r + 2)}(r - 8)\cancel{(r - 6)}} {-3\cancel{(r + 2)}\cancel{(r - 6)}(r + 3)} $ We are dividing by $r - 6$ , so $r - 6 \neq 0$ Therefore, $r \neq 6$ $q = \dfrac {r - 8} {-3(r + 3)} $ $ q = \dfrac{-(r - 8)}{3(r + 3)}; r \neq -2; r \neq 6 $